Alkylation of 1-naphthol in trifluoroethanol versus in DMSO

Question

I am supposed to work out which product is formed in which solvent when naphthalen-1-olate is treated with benzyl iodide.

DMSO is a polar aprotic solvent, as it does not possess an acidic hydrogen. So, the negative charge on oxygen will be concentrated as the anion will be poorly solvated. Hence, it should act as a hard nucleophile. Since alkyl halides such as BnI are soft electrophiles, I think that the substituent should preferentially add to the ring.

In 2,2,2-trifluoroethan-1-ol (TFE), I think the reaction would proceed via an S_N1 mechanism. Now, the carbocation is a hard electrophile, and thus O-alkylation might be observed.

But considering the better solvation of the negatively charged oxygen in TFE, it would also be more hindered, which suggests that C-alkylation might also be seen.

Which products are actually formed, and am I on the right track?

Answer 1

The 1-naphthoxide ion has negative charge on the oxygen and the 2 and 4 carbons. This can be seen in resonance structures such as this:

Since oxygen is significantly more electronegative than carbon we would expect the resonance structure with the negative charge on the oxygen to be the major contributor and so the charge density will be greatest on the oxygen. Therefore in DMSO, where solvation of the anion is poor, we would expect the reaction to proceed through the oxygen nucleophile with an $\mathrm{S_N2}$ mechanism due to the primary nature of the electrophile.

In trifluroethanol the oxygen will be strongly solvated by hydrogen bonding and so its nucleophilicity will be drastically reduced, partly due to steric hindrance of the solvating alcohol molecules. This allows substitution to occur elsewhere. The 4 position is probably preferred over the 2 position again due to the steric hindrance of the solvent coordinating to the adjacent oxygen.

EDIT: Thanks to @jerepierre for pointing this out in the comments: "One of the products (the carbon-alkylated one) requires breaking aromaticity, which is most likely going to be much slower than O-alkylation for either solvent system. A cursory search of SciFinder shows no examples of this reaction. I'm guessing that somebody made up this problem to test solvent effects in alkylation, but came up with a poor example. The best 'predict the products' questions are ones that have actually been performed."

Answer 2

These solvents should give identical products.

DMSO is considered the quintessential polar aprotic solvent. Given that the negative oxygen is going to fairly nucleophilic and that the second reactant has iodine on it (and that iodine is a leaving group of the Gods), my guess is that this reaction will proceed via SN2 mechanism to form the product they show for A in answer (A).

Similarly, fluorinated ethanol is very polar and protic. This reaction will proceed via SN1, with the iodine leaving and the carbocation stabilized by the phenyl ring it is attached to. I don't believe there should be significant carbocation rearrangement, since the stability of the product with the aromatic ring intact should outweigh the any added stability from the carbocation being on the ring, but don't quote me on that. So this will give a product identical to the one for DMSO. I would say that this mechanism can result in stereochemical differences, but the carbon being attacked by the oxide is not chiral.